Chemical Equilibrium - University of Massachusetts Bostonalpha.chem.umb.edu/chemistry/ch311/Week1.pdf · Properties of Umass Boston. Equilibrium. When the net reaction rate is equal

Properties of Umass Boston

Chemical Equilibrium


EquilibriumWhen the net reaction rate is equal to 0, the system becomes “dynamically equilibrium”.

Forward rate = Reverse rate

For reaction A + B C + D

r

f

rf

kk

BADCK

DCkBAkBADCK

==

=

=

]][[]][[

]][[]][[]][[]][[


Equilibrium Constants

For the reaction:aA +bB cC + dD

The Equilibrium Constant:

baa

dc

ba

dc

BPDCK

gasisAifBADCK

][][][

][][][][

=

=•A,B are the reactants and C,D products

•a,b,c,d are the stoichiometry coefficients

•[ ] stands for concentration

K tells us the position of equilibrium, K >1, more C and D


The Units • Standard State:

– For solutes: 1 M (mol/Liter).– For gases: 1 bar (1 bar = 105 Pa = 0.98682atm).– For solvent: pure material.

• The unit in the equilibrium calculation:– Concentration of solutes: M (mol/liter)– Gases: bars– Pure material: unit (1)


The unit of equilibrium constant

• Equilibrium Constant K is dimensionless

• Why?– The [A],[B],[C] and [D]

are ratios

)1(tan)(Pr

)1(tan)(][

barsstatedardsbarsessureP

MstatedardsMAofonconcentratA

a =

=


Forward and backward reactions

RF

dc

ba

R

ba

dc

F

K

K

KK

DCBAK

reactionverseBADCK

reactionforward

dDcCbBaAF

R

1][][][][

:Re][][][][

:

=

=

=

+⇔+


Adding reactions

• If multiple reactions are added, the new K is the product of the individual values

]][[]][[

]][[][

][]][[

_______________________

213

3

2

1

CHACHA

CHCH

HAAHKKK

KCHACHA

KCHCHKAHHA

+−+−+

+−

++

−+

=×==

+⇔+

+⇔+

+⇔


Equilibrium of Water

• H2O H+ + OH-

autoprotonlysis: self-ionization

Kw = [H+][OH-] = 1.0x10-14

• Where [H2O]?• In the aqueous solution, at 25oC, Kw ≡

1.0x10-14


Equilibrium and Thermodynamics• aA +bB cC + dD• ΔG (Free energy)=ΔH(enthalpy)-TΔS(entropy)• ΔG=ΔGo + RTlnQ where Q=[C]c[D]d/[A]a[B]b

Q: reaction quotientWhen, Q=1 standard state, ΔG=ΔGo

When, ΔG<0, favor forward reactionWhen, ΔG>0, favor reverse reactionWhen, ΔG=0, reaction at equilibrium, K=Q

SO ΔGo =- RTlnK

K=e-ΔGo/RT


Le Châtelier’s Principal

• When the equilibrium is disturb, the position of the reaction must shift to the direction which can partially off-set the disruption and try to bring the system back to equilibrium.


Adding reactant or product

For Reaction A B at equilibriumK=[B]/[A]

• If more A is added: In order to off-set the disruption, the reaction has to move forward to reduce amount of A.

• If more B is added: the reaction has to move backward to reduce amount of B.

• Does K change? Why?• NO, because the K is a state function.


Temperature change

• For endothermic reaction, ΔHo>0, K increase when T increase. (heat+A B)

• For exothermic reaction, ΔHo<0, K decrease when T increase. (A B+heat)

RSRTH

RSRTH

RTSTHRTG

oo

oo

ooo

ee

e

eeK

//

)//(

/)(/

ΔΔ−

Δ+Δ−

Δ−Δ−Δ−

=

=

==


Solubility

Dissolve

Precipitate

•When dissolution rate > precipitation rate, solid dissolves.

•When dissolution rate = precipitation rate, solution become saturated (under the condition of sufficient amount of solid), the equilibrium is reached.

•The concentration of the solid = 1, so forA (s) B + C; K = [B][C] = Ksp, when the solution becomes saturated, Ksp is reached.


Common Ion Effect• A (s) B + C

when the Ksp is reached, according to Le Châtelier’s principal, adding more B or C can precipitate more A.


Example• Hg2Cl2 dissolves in water

Hg2Cl2 (s) Hg22+ + 2Cl-

initial solid 0 0equ. Solid x 2x

Ksp = x*(2x)2 =1.2x10-18 → x=6.7x10-7M[Hg2

2+ ]=6.7x10-7M; [Cl -]=1.3x10-6 M• Adding 0.03 M [Cl] into saturated solution

Hg2Cl2 (s) Hg22+ + 2Cl-

initial solid 6.7x10-7 1.3x10-6

equ. Solid x 2x+0.03

Ksp = x*(2x+0.03)2=1.2x10-18 → x=1.3x10-15M[Hg2

2+ ]=1.3x10-15M; [Cl -]=0.03 M (2x is too small)

• If more Hg2Cl2 solid is added, according to Le Châtelier’s principal, Hg2

2+ + 2Cl- concentration should be higher, Is that right? Why?


Complex Formation• Adding I- into Pb2+ solution, PbI2 becomes precipitated:

PbI2(s) ↔ Pb 2+ + 2I- Ksp = 7.9*10-9 = [Pb2+][I-]2• Adding more I- will make PbI2(s) become dissolved. Because complex

formation:

– Pb 2++ I -↔ PbI+ (aq) K1 = 100 =[PbI+(aq)]/[Pb2+][I-] – Pb2+ +2I-↔ PbI2(aq) K2 =1400 = PbI2(aq)]/[Pb2+][I-]2– Pb2++ 3I-↔ PbI-3(aq) K3 = 8300 = [PbI-3(aq)]/[Pb2+][I-]3– Pb2++ 4I-↔ PbI2-

4 (aq) K4 = 30000 = [PbI2-4(aq)]/[Pb2+][I-]4

• [Pb2+] is the same in all six of these equilibrium expressions. There can only be one [Pb2+] and one [I-] concentration

• Define soluble lead – any soluble species containing lead contributes to the total concentration of soluble lead.Soluble lead = [Pb2+] + [PbI+(aq)] + [PbI2(aq)] + [PbI3-(aq)] + [PbI42-(aq

9/12/2011 Properties of Umass Boston

Acid and Base


• Lewis acid and baseLewis acid: orbit to accept electron pairLewis base: a pair of electron to be donated

• Protic acid and baseAcid: proton donor HClBase: proton acceptor NaOH

• Conjugate acid and base are related by the gain or loss of one proton


Acid – base reaction:

HCl (acid) + NaOH (base)→ NaCl (salt) + H2O

CH3COOH +CH3NH2 CH3NH3+ + CH3COO-


pH

• Definition: pH=-log[H+]; acidity of solution• For pure water: Kw=[H+][OH-]=1.0x10 -14

(25oC)So [H+]=[OH-]=10-7; pH=-log(10-7)=7

• In acidic solution, e.g. [H+]=10-4, pH=4• In alkaline solution, e.g. [OH-]=10-4

[H+]=Kw/[OH-]=10-14/10-4=10-10; pH=10• pH > 7 alkaline; pH < 7 acidic

In aqueous solution, [H+][OH-]≡10-14 at 25 oC


Strength of acid and base

• Strong acid or base: it can be completely dissociated in the solution, e.g. in water

HCl → H+ + Cl-; KOH → K+ + OH-

• Week acid or base: it can only be partially dissociated in the solution e.g.

CH3COOH CH3COO- + H+

Ka = [CH3COO-][H+]/[CH3COOH](acid dissociation constant)



• Dissociation Vs. Dissolution• Hydrogen Halides HCl, HBr and HI are

strong acid.• HF is specially

HF → H+ + F- F-….H+ (ion pair)



• Solution consists of:– Salt formed by strong acid and strong base is

neutral, e.g. NaCl (HCl/NaOH)– Salt formed by strong acid and weak base is acidic

e.g. NH4Cl (NH4OH/HCl)NH4Cl + H2O NH4OH + H+ + Cl-

– Salt formed by weak acid and strong base is basic, e.g. CH3COONa (CH3COOH/NaOH)CH3COONa + H2O CH3COOH + Na+ + OH-


At low concentration systematic treatment is needed

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Chemical Equilibrium - University of Massachusetts Bostonalpha.chem.umb.edu/chemistry/ch311/Week1.pdf · Properties of Umass Boston. Equilibrium. When the net reaction rate is equal